# Count duplicates in a given linked list

Given a linked list. The task is to count the number of duplicate nodes in the linked list.
Examples:

Input: 5 -> 7 -> 5 -> 1 -> 7 -> NULL
Output: 2

Input: 5 -> 7 -> 8 -> 7 -> 1 -> NULL
Output: 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Simple Approach: We traverse the whole linked list. For each node we check in the remaining list whether the duplicate node exists or not. If it does then we increment the count.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Representation of node ` `struct` `Node { ` `    ``int` `data; ` `    ``Node* next; ` `}; ` ` `  ` `  `// Function to insert a node at the beginning ` `void` `insert(Node** head, ``int` `item) ` `{ ` `    ``Node* temp = ``new` `Node(); ` `    ``temp->data = item; ` `    ``temp->next = *head; ` `    ``*head = temp; ` `} ` ` `  `// Function to count the number of ` `// duplicate nodes in the linked list ` `int` `countNode(Node* head) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``while` `(head->next != NULL) { ` ` `  `        ``// Starting from the next node ` `        ``Node *ptr = head->next; ` `        ``while` `(ptr != NULL) { ` ` `  `            ``// If some duplicate node is found ` `            ``if` `(head->data == ptr->data) { ` `                ``count++; ` `                ``break``; ` `            ``} ` `            ``ptr = ptr->next; ` `        ``} ` ` `  `        ``head = head->next; ` `    ``} ` ` `  `    ``// Return the count of duplicate nodes ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``Node* head = NULL; ` `    ``insert(&head, 5); ` `    ``insert(&head, 7); ` `    ``insert(&head, 5); ` `    ``insert(&head, 1); ` `    ``insert(&head, 7); ` ` `  `    ``cout << countNode(head); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG ` `{ ` `     `  `// Representation of node  ` `static` `class` `Node  ` `{  ` `    ``int` `data;  ` `    ``Node next;  ` `};  ` ` `  ` `  `// Function to insert a node at the beginning  ` `static` `Node insert(Node head, ``int` `item)  ` `{  ` `    ``Node temp = ``new` `Node();  ` `    ``temp.data = item;  ` `    ``temp.next = head;  ` `    ``head = temp; ` `    ``return` `head; ` `}  ` ` `  `// Function to count the number of  ` `// duplicate nodes in the linked list  ` `static` `int` `countNode(Node head)  ` `{  ` `    ``int` `count = ``0``;  ` ` `  `    ``while` `(head.next != ``null``)  ` `    ``{  ` ` `  `        ``// Starting from the next node  ` `        ``Node ptr = head.next;  ` `        ``while` `(ptr != ``null``)  ` `        ``{  ` ` `  `            ``// If some duplicate node is found  ` `            ``if` `(head.data == ptr.data)  ` `            ``{  ` `                ``count++;  ` `                ``break``;  ` `            ``}  ` `            ``ptr = ptr.next;  ` `        ``}  ` ` `  `        ``head = head.next;  ` `    ``}  ` ` `  `    ``// Return the count of duplicate nodes  ` `    ``return` `count;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String args[]) ` `{  ` `    ``Node head = ``null``;  ` `    ``head = insert(head, ``5``);  ` `    ``head = insert(head, ``7``);  ` `    ``head = insert(head, ``5``);  ` `    ``head = insert(head, ``1``);  ` `    ``head = insert(head, ``7``);  ` ` `  `    ``System.out.println( countNode(head));  ` `} ` `}  ` ` `  `// This code is contributed by Arnab Kundu `

## Python3

 `# Python3 implementation of the approach ` `import` `math ` ` `  `# Representation of node ` `class` `Node:  ` `    ``def` `__init__(``self``, data):  ` `        ``self``.data ``=` `data  ` `        ``self``.``next` `=` `None` ` `  `# Function to push a node at the beginning ` `def` `push(head, item):  ` `    ``temp ``=` `Node(item);  ` `    ``temp.data ``=` `item;  ` `    ``temp.``next` `=` `head;  ` `    ``head ``=` `temp; ` `    ``return` `head; ` `         `  `# Function to count the number of ` `# duplicate nodes in the linked list ` `def` `countNode(head): ` `    ``count ``=` `0` `    ``while` `(head.``next` `!``=` `None``): ` `         `  `        ``# print(1) ` `        ``# Starting from the next node ` `        ``ptr ``=` `head.``next` `        ``while` `(ptr !``=` `None``): ` `             `  `            ``# print(2) ` `            ``# If some duplicate node is found ` `            ``if` `(head.data ``=``=` `ptr.data): ` `                ``count ``=` `count ``+` `1` `                ``break` `            ``ptr ``=` `ptr.``next` `        ``head ``=` `head.``next` `         `  `    ``# Return the count of duplicate nodes ` `    ``return` `count ` ` `  `# Driver code ` `if` `__name__``=``=``'__main__'``:  ` ` `  `    ``head ``=` `None``; ` `    ``head ``=` `push(head, ``5``) ` `    ``head ``=` `push(head, ``7``) ` `    ``head ``=` `push(head, ``5``) ` `    ``head ``=` `push(head, ``1``) ` `    ``head ``=` `push(head, ``7``) ` `    ``print``(countNode(head)) ` ` `  `# This code is contributed by Srathore `

## C#

 `// C# implementation of the approach  ` `using` `System; ` `     `  `class` `GFG ` `{ ` `     `  `// Representation of node  ` `public` `class` `Node  ` `{  ` `    ``public` `int` `data;  ` `    ``public` `Node next;  ` `};  ` ` `  ` `  `// Function to insert a node at the beginning  ` `static` `Node insert(Node head, ``int` `item)  ` `{  ` `    ``Node temp = ``new` `Node();  ` `    ``temp.data = item;  ` `    ``temp.next = head;  ` `    ``head = temp; ` `    ``return` `head; ` `}  ` ` `  `// Function to count the number of  ` `// duplicate nodes in the linked list  ` `static` `int` `countNode(Node head)  ` `{  ` `    ``int` `count = 0;  ` ` `  `    ``while` `(head.next != ``null``)  ` `    ``{  ` ` `  `        ``// Starting from the next node  ` `        ``Node ptr = head.next;  ` `        ``while` `(ptr != ``null``)  ` `        ``{  ` ` `  `            ``// If some duplicate node is found  ` `            ``if` `(head.data == ptr.data)  ` `            ``{  ` `                ``count++;  ` `                ``break``;  ` `            ``}  ` `            ``ptr = ptr.next;  ` `        ``}  ` ` `  `        ``head = head.next;  ` `    ``}  ` ` `  `    ``// Return the count of duplicate nodes  ` `    ``return` `count;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main(String []args) ` `{  ` `    ``Node head = ``null``;  ` `    ``head = insert(head, 5);  ` `    ``head = insert(head, 7);  ` `    ``head = insert(head, 5);  ` `    ``head = insert(head, 1);  ` `    ``head = insert(head, 7);  ` ` `  `    ``Console.WriteLine( countNode(head));  ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```2
```

Time Complexity : O(n*n)

Efficient Approach: The idea is to use hashing

## C++

 `// C++ implementation of the approach ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Representation of node ` `struct` `Node { ` `    ``int` `data; ` `    ``Node* next; ` `}; ` ` `  `// Function to insert a node at the beginning ` `void` `insert(Node** head, ``int` `item) ` `{ ` `    ``Node* temp = ``new` `Node(); ` `    ``temp->data = item; ` `    ``temp->next = *head; ` `    ``*head = temp; ` `} ` ` `  `// Function to count the number of ` `// duplicate nodes in the linked list ` `int` `countNode(Node* head) ` `{ ` `    ``if` `(head == NULL) ` `       ``return` `0;; ` ` `  `    ``// Create a hash table insert head ` `    ``unordered_set<``int``> s; ` `    ``s.insert(head->data); ` ` `  `    ``// Traverse through remaining nodes     ` `    ``int` `count = 0; ` `    ``for` `(Node *curr=head->next; curr != NULL; curr=curr->next) { ` `        ``if` `(s.find(curr->data) != s.end()) ` `             ``count++;  ` ` `  `        ``s.insert(curr->data); ` `    ``} ` ` `  `    ``// Return the count of duplicate nodes ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``Node* head = NULL; ` `    ``insert(&head, 5); ` `    ``insert(&head, 7); ` `    ``insert(&head, 5); ` `    ``insert(&head, 1); ` `    ``insert(&head, 7); ` ` `  `    ``cout << countNode(head); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.HashSet; ` ` `  `class` `GFG  ` `{ ` ` `  `// Representation of node ` `static` `class` `Node ` `{ ` `    ``int` `data; ` `    ``Node next; ` `}; ` `static` `Node head; ` ` `  `// Function to insert a node at the beginning ` `static` `void` `insert(Node ref_head, ``int` `item) ` `{ ` `    ``Node temp = ``new` `Node(); ` `    ``temp.data = item; ` `    ``temp.next = ref_head; ` `    ``head = temp; ` `     `  `} ` ` `  `// Function to count the number of ` `// duplicate nodes in the linked list ` `static` `int` `countNode(Node head) ` `{ ` `    ``if` `(head == ``null``) ` `    ``return` `0``;; ` ` `  `    ``// Create a hash table insert head ` `    ``HashSets = ``new` `HashSet<>(); ` `    ``s.add(head.data); ` ` `  `    ``// Traverse through remaining nodes  ` `    ``int` `count = ``0``; ` `    ``for` `(Node curr=head.next; curr != ``null``; curr=curr.next)  ` `    ``{ ` `        ``if` `(s.contains(curr.data)) ` `            ``count++;  ` ` `  `        ``s.add(curr.data); ` `    ``} ` ` `  `    ``// Return the count of duplicate nodes ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` ` `  `    ``insert(head, ``5``); ` `    ``insert(head, ``7``); ` `    ``insert(head, ``5``); ` `    ``insert(head, ``1``); ` `    ``insert(head, ``7``); ` ` `  `    ``System.out.println(countNode(head)); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic; ` `     `  `class` `GFG  ` `{ ` ` `  `// Representation of node ` `public` `class` `Node ` `{ ` `    ``public` `int` `data; ` `    ``public` `Node next; ` `}; ` `static` `Node head; ` ` `  `// Function to insert a node at the beginning ` `static` `void` `insert(Node ref_head, ``int` `item) ` `{ ` `    ``Node temp = ``new` `Node(); ` `    ``temp.data = item; ` `    ``temp.next = ref_head; ` `    ``head = temp; ` `     `  `} ` ` `  `// Function to count the number of ` `// duplicate nodes in the linked list ` `static` `int` `countNode(Node head) ` `{ ` `    ``if` `(head == ``null``) ` `    ``return` `0;; ` ` `  `    ``// Create a hash table insert head ` `    ``HashSet<``int``>s = ``new` `HashSet<``int``>(); ` `    ``s.Add(head.data); ` ` `  `    ``// Traverse through remaining nodes  ` `    ``int` `count = 0; ` `    ``for` `(Node curr=head.next; curr != ``null``; curr=curr.next)  ` `    ``{ ` `        ``if` `(s.Contains(curr.data)) ` `            ``count++;  ` ` `  `        ``s.Add(curr.data); ` `    ``} ` ` `  `    ``// Return the count of duplicate nodes ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args)  ` `{ ` ` `  `    ``insert(head, 5); ` `    ``insert(head, 7); ` `    ``insert(head, 5); ` `    ``insert(head, 1); ` `    ``insert(head, 7); ` ` `  `    ``Console.WriteLine(countNode(head)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```2
```

Time Complexity : O(n)

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.